## Giải bài 5 với những yêu cầu như trên

Question

Giải bài 5 với những yêu cầu như trên

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12 months 2020-11-28T20:17:14+00:00 1 Answers 45 views 0

## Answers ( )

1. Giải thích các bước giải:

Ta có:

$$\begin{array}{l} a,\\ DKXD:\,\,\,\left\{ \begin{array}{l} x \ge 0\\ x – 1 \ne 0\\ x + 2\sqrt x + 1 \ne 0 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} x \ge 0\\ x \ne 1 \end{array} \right.\\ b,\\ P = \left( {\dfrac{{\sqrt x – 2}}{{x – 1}} – \dfrac{{\sqrt x + 2}}{{x + 2\sqrt x + 1}}} \right).\dfrac{{{{\left( {1 – x} \right)}^2}}}{2}\\ = \left( {\dfrac{{\sqrt x – 2}}{{\left( {\sqrt x – 1} \right)\left( {\sqrt x + 1} \right)}} – \dfrac{{\sqrt x + 2}}{{{{\left( {\sqrt x + 1} \right)}^2}}}} \right).\dfrac{{{{\left( {x – 1} \right)}^2}}}{2}\\ = \dfrac{{\left( {\sqrt x – 2} \right)\left( {\sqrt x + 1} \right) – \left( {\sqrt x + 2} \right)\left( {\sqrt x – 1} \right)}}{{\left( {\sqrt x – 1} \right){{\left( {\sqrt x + 1} \right)}^2}}}.\dfrac{{{{\left( {\sqrt x – 1} \right)}^2}.{{\left( {\sqrt x + 1} \right)}^2}}}{2}\\ = \dfrac{{\left( {x – \sqrt x – 2} \right) – \left( {x + \sqrt x – 2} \right)}}{{\left( {\sqrt x – 1} \right){{\left( {\sqrt x + 1} \right)}^2}}}.\dfrac{{{{\left( {\sqrt x – 1} \right)}^2}.{{\left( {\sqrt x + 1} \right)}^2}}}{2}\\ = \dfrac{{ – 2\sqrt x }}{1}.\dfrac{{\sqrt x – 1}}{2}\\ = – \sqrt x \left( {\sqrt x – 1} \right)\\ c,\\ P = – \sqrt x \left( {\sqrt x – 1} \right)\\ = – \left( {x – \sqrt x } \right)\\ = – \left( {x – \sqrt x + \dfrac{1}{4}} \right) + \dfrac{1}{4}\\ = \dfrac{1}{4} – {\left( {\sqrt x – \dfrac{1}{2}} \right)^2} \le \dfrac{1}{4},\,\,\,\forall x \ge 0,\,\,x \ne 1\\ \Rightarrow {P_{\max }} = \dfrac{1}{4} \Leftrightarrow {\left( {\sqrt x – \dfrac{1}{2}} \right)^2} = 0 \Leftrightarrow \sqrt x = \dfrac{1}{2} \Leftrightarrow x = \dfrac{1}{4}\\ d,\\ 0 < x < 1 \Rightarrow 0 < \sqrt x < 1\\ \Rightarrow \sqrt x – 1 < 0\\ \Rightarrow \sqrt x .\left( {\sqrt x – 1} \right) < 0\\ \Leftrightarrow – \sqrt x .\left( {\sqrt x – 1} \right) > 0\\ \Leftrightarrow P > 0 \end{array}$$